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1. Metals in group 1 have (+1) ox #, group 2 metals (+2) 2. Any single “Pure” element = 0 3. Hydrogen is always (+1) except in metal hydride (-1) LiH.

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Presentation on theme: "1. Metals in group 1 have (+1) ox #, group 2 metals (+2) 2. Any single “Pure” element = 0 3. Hydrogen is always (+1) except in metal hydride (-1) LiH."— Presentation transcript:

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2 1. Metals in group 1 have (+1) ox #, group 2 metals (+2) 2. Any single “Pure” element = 0 3. Hydrogen is always (+1) except in metal hydride (-1) LiH 4. Oxygen is always (-2) exceptions: With flourine (flouide) +2 OF 2 In Peroxides (-1) H 2 O 2 5. The sum of all oxidation #’s must = 0 6. The sum of all Polyatomic ions must equal the charge of that ion

3  Binary Compounds _____ 1. Start with the Non Metal 2. Finish with the Metal HCl 3. Sum up must = 0 _______ MgCl

4  Ternary Compounds _____ 1. Start with the Non Metal (Oxygen) H 2 SO 4 2. Go to the Metal (H) 3. finish up in the middle 4. Sum up must = 0 __________ Mg(NO 3 ) 2

5 1. Oxidation Reduction Loss of Electrons Half Reactions Mg 0  Mg +2 + 2e - (product) The gain of Electrons Half Reaction 2e - + Mg +2  Mg 0 (reactant) Causes the Reduction of the other elements Acts as a REDUCING AGENT (R.A.) Causes the other species to be Oxidized. Acts as the Oxidation Agent (O.A.)

6 Steps (Now this is Doc’s Method! …..Capisco?) 1. Assign the Ox #’s 2. Record the changes 3. Record e- loss / e- gain 4. Determine the species that is oxidized (RA) and reduced (OA) 5. Balance if unequal Ca + Cl 2  CaCl 2 All Redox Reactions must demonstrate conservation of both Mass and Charge ***HHH___

7 AlCu

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15 2H 2 O  2H 2 + O 2

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17 2NaCl  2Na + Cl 2


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